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Question

Two 29Cu64 nuclei almost touch each other. The electronics repulsive energy of the system will be (R0=1 fm)

A
788 MeV
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B
7.88 MeV
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C
126.15 MeV
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D
0.788 MeV
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Solution

The correct option is C 126.15 MeV
Radius of each nucleus,

R=R0(A)1/3=1.2(64)1/3=4.8 fm=4.8×1015 m

Distance between two nuclei (r)=2R=2×4.8×1015 m

So, potential energy

U=kq2r=9×109×(1.6×1019×29)22×4.8×1015

Potential energy in (eV),

U=kq2rqelectron=9×109×(1.6×1019×29)22×4.8×1015×1.6×1019

=126.15×106 eV=126.15 MeV

Hence, (C) is the correct answer.

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